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2 years ago

What is the gravitational field strength of a 5 kg object that weighs 25 N?

Physics

2 answers:

Over [174]2 years ago

8 0

Answer:

gravitation field strength: 5 N/kg

Explanation:

Weight(N) = mass(kg) * gravitational field strength(N/kg)

Here given the mass: 5 kg and weight: 25 N

make sure the units are proper and make change only if different units visible. Like mass given in grams and covert it into kg. *must* and sometimes weight given in kilo-newton covert it into Newtons. *must*

using the formula:

5 * gravitational field strength = 25

gravitational field strength = 25/5

gravitational field strength = 5 N/kg

solniwko [45]2 years ago

7 0

Answer:

ExplanaThis is a way of measuring how much gravity there is. The formula is: weight/mass = gravitational field strength.

Gravitational field strength = Weight/mass unit is N/kg

Weight = mass x gravitational field strength unit is N

On Earth the gravitational field strength is 10 N/kg. Other planets have different gravitational field strengths. The Moon has a gravitational field strength of 1.6 N/kg. You might have seen films of astronauts leaping high on the moon.

Here on Earth, if I jump I am pulled back to ground by gravity. What is my weight? My mass is 80kg and if we multiply by gravitational field strength (10N/kg) - my weight is 800N. Now if I go to the moon, my mass will be the same, 80kg. We multiply that by the moon's gravitational field strength, which is 1.6 N/ kg. That means my weight on the moon is 128N. So I have different weights on the Earth and on the Moon. That's why astronauts can jump high into the air on the moon - they're lighter up there.

Jupiter is a very large planet with strong gravitational field strength of 25 N/ kg. My body is 80kg. If I go to Jupiter my weight is going to be 25 x 80 = 2,000 N. That means I wouldn't be able to get off the ground or stand up straight! I would probably be lying down all the time there. So weight varies depending on which planet you are on. You can find out more yourself by looking up tables of weight on different planets.tion:

pls brainlieste

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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

3 years ago

A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of

Paha777 [63]

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

$Workdone = force * distance$

Making "distance" the subject of formula, we have;

$Distance = \frac {workdone}{force}$

Substituting into the equation, we have;

$Distance = \frac {195}{56.7}$

Distance = 3.4 meters.

8 0

3 years ago

A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c

Artist 52 [7]

Answer:

Approximately $25\; {\rm N}$ (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

When the $6.0\; {\rm cm}$ -spring in this question is stretched to $10\; {\rm cm}$ , the displacement is $x = (10\; {\rm cm} - 6.0\; {\rm cm})$ .
Likewise, if this spring is stretched to $20\; {\rm cm}$ , the displacement would be $(20\; {\rm cm} - 6\; {\rm cm})$ .

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of $F_{\text{a}}$ displaces this spring by $x_{\text{a}}$ , while a force of $F_{\text{b}}$ displaces this spring by $x_{\text{b}}$ , then:

$\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}$ .

In this question, it is given that a force of $F_{\text{a}} = 7.0 \; {\rm N}$ would stretch this spring by $x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm})$ . Thus, the force $F_{\text{b}}$ required to stretch this spring by $x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm})$ would satisfy:

$\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}$ .

Rearrange and solve for $F_{\text{b}}$ :

$\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}$ .